Some Remarks on Bielliptic and Trigonal Curves

Home , Hyperelliptic curve

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Maclachlan’s theorem says that hyperelliptic curves cannot have automorphisms of certain types, for example ﬁxed-point-free of prime order p 3. In the sequel we want to generalize this to other types of curves. ≥ On the other hand, it is clear that the curve Y = X/G in Theorem 1.1 must also be hyperelliptic. So another interpretation is that certain coverings X of a hyperelliptic curve Y cannot be hyperelliptic. In this form we also want to generalize it, keeping the curve Y hyperelliptic, but taking other types of curves X. One generalization of a hyperelliptic involution is the notion of a b-hyperelliptic involution, that is, an involution such that the genus of the quotient curve is b. We concentrate on the case b = 1; then the involution and the curve are called bielliptic. Some of the literature also uses the expression elliptic-hyperelliptic for this. Another generalization of X being hyperelliptic is the existence of an n-gonal map. This is a surjective morphism of degree n from X to a projective line. Note that for n > 2 such a morphism is not necessarily Galois, i.e. does not necessarily come from an automorphism of X. This makes it much more diﬃcult to establish or exclude the existence of such a map. We are mainly interested in the case n = 3; then the map and the curve are called trigonal. Throughout the paper we will freely use the following easy consequences of the Castelnuovo inequality:

The hyperelliptic involution is unique. • On a bielliptic curve of genus g 6 the bielliptic involution is unique. • ≥ A curve of genus g 5 has at most one trigonal map. • ≥ A hyperelliptic curve of genus g 3 cannot be trigonal. • ≥

2 A hyperelliptic curve of genus g 4 cannot be bielliptic. • ≥ Finally, we’ll need the following results.

Theorem 1.2. Let Y be a hyperelliptic curve of genus h, and let X Y be a → smooth, cyclic covering of degree n.

(a) (Bujalance) [B, Theorem] If n =2, then X is b-hyperelliptic for some b with with b h−1 . ≤ ⌊ 2 ⌋ In particular, a smooth degree 2 cover of a genus 2 curve is a genus 3 curve that is hyperelliptic (and bielliptic).

(b) (Accola) [A, Lemma 2] If n is odd, then the hyperelliptic involution of Y lifts (n−1)(h−1) to n diﬀerent involutions on X that are all b-hyperelliptic where b = 2 . Together with the automorphism of order n from Gal(X/Y ) they generate a dihedral group Dn of order 2n. In particular, a smooth degree 3 Galois cover of a genus 2 curve is a genus 4 curve that has at least 3 bielliptic involutions.

2. Bielliptic curves We start with a quick corollary to the known results from the previous section.

Corollary 2.1. Let X be a curve of genus 5. If X has an automorphism σ of order 4 such that σ has no ﬁxed points, then X is bielliptic.

Proof. As X/ σ2 is an unramified degree 2 cover of X/ σ , the genus of X/ σ2 h i h i h i can only be 1 (if σ2 has fixed points) or 3. In the first case we are done. In the second case we get a chain X X/ σ2 X/ σ of smooth coverings of degree 2 with genera 5, 3, and 2. By Theorem→ h i 1.2 → (a) theh i genus 3 curve is hyperelliptic, and the genus 5 curve is hyperelliptic or bielliptic. But it cannot be hyperelliptic by Theorem 1.1.

Ultimately Theorem 1.2 and Corollary 2.1 are based on lifting the hyperelliptic involution to an unramified cover. Our next result requires lifting the hyperelliptic involution to a ramified cover, which is a much more tricky problem. The paper [CT] discusses conditions under which this is possible. Compared to our previous statements we change the names of the curves and their automorphisms so that they fit precisely with those in the somewhat technical conditions in [CT].

Proposition 2.2. Let Y be a non-hyperelliptic curve of genus 4. If Y has an automorphism τ of order 4, then Y is bielliptic.

3 Proof. Since Y is non-hyperelliptic, the curve X = Y/ τ 2 can only have genus 1 h i or 2. In the first case we are done. In the second case τ 2 has exactly 2 fixed points P and Q on Y . From the Hurwitz formula we see that Y/ τ has genus 1 and τ also fixes P and Q. So τ, the involution h i induced by τ on X is different from σ, the hyperelliptic involutione of X. As σ commutes with τ, it acts on its fixed points Pe and Qe, the images of P and Q on X. Moreover, differente involutions must have disjoint fixed points. Hence σ(Pe) = Qe. Since the Weierstrass points of X are exactly the fixed points of σ, we see that Pe and Qe = σ(Pe), the only two points on X that ramify in Y , are not Weierstrass points of X. So our constellation satifies the conditions of [CT, Theorem 2.1], which allows us to conclude that σ lifts to an automorphism of Y . Actually, σ cannot lift to an automorphism ψ of order 4, because then we would have ψ2 = τ 2 and as before X/ σ = Y/ ψ would have genus 1. So σ lifts to an involution ψ on Y that commutesh i with τ 2h. Theni the other two intermediate curves between Y and the genus zero curve Y/ τ 2, ψ have genera 2 and 0 (which would h i contradict non-hyperellipticity of Y ) or 1 and 1, which implies that in this case Y actually has at least 2 bielliptic involutions.

Here is an analogue of Maclachlan’s result for bielliptic curves.

Proposition 2.3. Let Y be a hyperelliptic curve of genus at least 4 and let X Y be a smooth Galois cover with Galois group G such that X is bielliptic. Then →

(a) At least half of the elements of G are involutions.

(b) If G is abelian, then G has exponent 2 and G divides 8. | | Proof. (a) Since g(X) 6, the bielliptic involution τ is unique and hence commutes with G. As Y cannot be≥ bielliptic, τ induces the hyperelliptic involution on Y . Counting ramified points in the covering X X/ τ,G as in the proof of the second claim of Theorem 1.1, one sees that G must→ haveh at leasti G /2 involutions. (b) If G is abelian, part (a) implies that G| must| be 2-elementary abelian. In general, G is isomorphic to a subgroup of Aut(X/ τ ). But the biggest 2- elementary abelian group contained in the automorphismh i group of a genus one curve is Z/2Z Z/2Z Z/2Z. ⊕ ⊕ If in Proposition 2.3 we put the condition that the curve Y is bielliptic instead of hyperelliptic, then the number of fixed points of the bielliptic involution fits with the Hurwitz formula for smooth covers, and the smoothness of the cover fits with the fact that a surjective map between two curves of genus 1 is unramified. In short, the method of proof does not seem to imply any restrictions. On the other hand, for hyperelliptic Y we can even say something about the non-smooth case. But we need a little preparation.

4 Lemma 2.4. Let X be a bielliptic curve of genus g 6 with an automorphism σ of prime order p 3. ≥ ≥ (a) If σ has fixed points, then necessarily p =3. (b) If σ has no fixed points, then the curve X/ σ is also bielliptic. h i Proof. As g 6, the bielliptic involution τ is unique. So τ and σ commute. For (a) we≥ use that σ induces an automorphism σ of order p with fixed points on the genus one curve X/ τ , which is only possiblee for p = 3. For (b) note that any fixedh i point of σ on X/ τ would lift to one or two fixed points of σ on X. So σ is fixed-point-freee and henceh i X/ τ, σ has genus 1. This h i means that τ induces ae bielliptic involution on X/ σ . h i Proposition 2.5. Let Y be a hyperelliptic curve of genus at least 4. If X Y is → a Galois cover of degree relatively prime to 6, then X cannot be bielliptic.

Proof. Assume that X is bielliptic. As g(X) 6, the bielliptic involution τ is unique and lies in the center of Aut(X). Moreover,≥ because of g(Y ) 4 the Galois group G of the covering X Y cannot contain τ. So G induces an isomorphic≥ group → of automorphisms on X/ τ . As ﬁnite subgroups of automorphisms on a genus 1 curve are solvable, G is solvable.h i (Alternatively, we could argue here with the highly non-trivial fact that groups of odd order are solvable.) So there exists a chain of coverings X = X X Xs = Y 1 → 2 →···→ such that each covering Xi Xi is Galois of prime degree. By Lemma 2.4 (a) → +1 the biellipticity of Xi implies that the covering Xi Xi is unramiﬁed, and by → +1 Lemma 2.4 (b) this implies that Xi+1 is bielliptic, too. By iteration we get that Y is bielliptic, which cannot be for a hyperelliptic curve of genus bigger than 3.

Remark 2.6. With practically the same proof Proposition 2.5 holds under slightly more general conditions, for example if the degree of the Galois cover is odd and none of the ramiﬁcation indices is divisible by 3.

A bielliptic curve X of genus g(X) 5 can have more than one bielliptic involution. We recall some curves that reach≤ the maximally possible number of bielliptic involutions for the given genus.

The maximal possible number of automorphisms on a curve of genus 3 is 168. The unique genus 3 curve that realizes this bound is the modular curve X(7). Its automorphism group is the simple group P SL2(F7). This group is also isomorphic to GL3(F2). The projective model of this curve is x3y + y3z + z3x =0,

5 which is why it is also called the Klein quartic. Since its automorphism group is simple, this curve cannot be hyperelliptic. So by Theorem 1.2 (b) every involution must be bielliptic. Thus it has 21 bielliptic involutions, the maximum possible for a curve of genus 3.

By [CD, Corollary 6.9] a curve of genus 4 can have at most 10 bielliptic involutions, and there is exactly one curve of genus 4 with 10 bielliptic involutions. It is called Bring’s curve, and it is isomorphic to the modular curve X1(5, 10). Its automorphism group is isomorphic to the symmetric group S5. Actually, 120 is also the maximum number of automorphisms for a curve of genus 4.

By [KMV] a curve of genus 5 can have 0, 1, 2, 3 or 5 bielliptic involutions, and the genus 5 curves with 5 biellliptic involutions form a 2-dimensional family. They are called Humbert curves. Their automorphism groups have order 160. This does not quite reach the maximally possible number of automorphisms on a curve of genus 5, which is 192.

For the proof of the next result we have to introduce yet another type of curve with an obvious automorphism of order p, namely for each prime p 5 the Lef- schetz curve ≥ yp = x(x 1) − p−1 Z Z of genus 2 . Its full automorphism group is isomorphic to /2p . The unique involution x 1 x is hyperelliptic, and hence the curve is not bielliptic. 7→ − In [JKS2] we showed that there is only one bielliptic curve of genus 4 with an automorphism of order 5, namely Bring’s curve. Now we want to generalize this.

Theorem 2.7. Let X be a bielliptic curve. Assume that X has an automorphism σ of prime order p 5 with ﬁxed points. Then one of the following holds: ≥ (a) g =5, p =5 and X is a Humbert curve;

(b) g =4, p =5 and X is Bring’s curve;

Proof. By Lemma 2.4 (a) we must have g 5. Moreover, σ acts by conjugation ≤ on the bielliptic involutions. If it commutes with a bielliptic involution, it induces an automorphism of order p with ﬁxed points on the elliptic quotient curve, which is impossible. Hence the number of biellitic involutions must be divisible by p. If g = 5, the maximal number of bielliptic involutions is 5 [KMV], so necessarily p = and X has 5 bielliptic involutions, i.e. X is a Humbert curve. If g = 4, the possible numbers of bielliptic involutions are 0, 1, 2, 3, 4, 6, 10 [CD, p.600]. So p = 5 and 10 bielliptic involutions, i.e. X is Bring’s curve.

6 If g = 3, the possible prime divisors of Aut(X) are 2, 3 and 7. By [RR, Theorem | | 1] there are exactly two curves of genus 3 with an automorphism of order 7. One is the Klein quartic, the other one is the Lefschetz curve (which is not bielliptic). If g = 2, the biggest possible prime divisor of Aut(X) is 5. By [RR, p.199] the | | only genus 2 curve with an automorphism of order 5 is the Lefschetz curve, which is not bielliptic.

3. Trigonal curves We give another generalization of Maclachlan’s theorem.

Theorem 3.1. Let p be an odd prime. Let X be a p-gonal curve of genus g > (p 1)2 and G Aut(X) a subgroup of fixed-point-free automorphisms. Then G divides− p. ≤ | | Proof. Since g > (p 1)2, by the Castelnuovo inequality the p-gonal map π : X Y is unique. So G induces− an isomorphic group of automorphisms on the genus→ zero curve Y . Then G acts without fixed points on the ramification points P of π on X. Consequently, they come in batches of size G with the same ramification index eP . | | Hence G divides P(eP 1). Thus by the Hurwitz formula G divides 2g 2+2p. | | − | | − On the other hand, as the covering X X/G is smooth, G divides 2g 2. Subtracting this, we see that G divides 2p.→ | | − | | So what is left is showing that every involution σ on X must have fixed points. Let P be one of the two fixed points of the induced involution σ on Y . Let P1,...,Pr e be the points on X lying above P and let e1,...,er be their ramification indices. Then σ acts on these points. If σ fixes no Pi, then they come in pairs with the same r ramification index. So p = Pi=1 ei would be even, a contradiction.

Theorem 3.1 applies in particular to trigonal curves of genus g 5. ≥ Proposition 3.2. Let Y be a hyperelliptic curve of genus h 3, and let X Y be a Galois cover. Then X cannot be trigonal. ≥ →

Proof. If X is trigonal, then, because of g(X) 5, the trigonal map X Z is unique. So G, the Galois group of X Y , induces≥ an isomorphic group of auto-→ → morphisms on Z. This implies the existence of a trigonal map Y Z/G. But by Castelnuovo Y cannot be trigonal. →

Corollary 3.3. Let X Y be a smooth Galois cover of degree n where Y is a hyperelliptic curve of genus→ h. Then X is trigonal if and only if n =3, h =2, and X has genus 4.

Proof. Assume that X is trigonal. If g(X) 5, then n = 3 by Theorem 3.1 ≥

7 and h = 2 by Proposition 3.2; but this contradicts g(X) 5. So for trigonal X we ≥ must have g(X) 4. Now the Hurwitz formula leaves only the possibilities g = 4, n = 3, h =2 or g≤= 3, n = 2, h = 2. But by Theorem 1.2 (a) the second possibility would imply the contradiction that X is hyperelliptic. Conversely, a curve of genus 4 is either hyperelliptic or trigonal. But by Theorem 1.1 a smooth Galois cover of degree 3 cannot be hyperelliptic.

Concerning involutions on trigonal curves we have the following result.

Lemma 3.4. Let X be a trigonal curve of genus g and σ an involution on X. (a) If g is odd, then σ has exactly 4 fixed points. (b) If g is even, then σ has 2 or 6 fixed points. Proof. If g 5, the trigonal map X Y is unique, and hence σ induces an ≥ → involution σ on Y . Since σ has exactly two fixed points on Y , σ can have at most 6 fixed pointse on X. By thee Hurwitz formula this means 2 or 6 fixed points if g is even, and 0 or 4 if g is odd. But 0 is excluded by Theorem 2.1. If g = 4, then a priori there could be 2, 6 or 10 fixed points. But 10 fixed points would mean that the involution is hyperelliptic, which is not possible for a trigonal curve of genus bigger than 2. For the same reason 8 fixed points are not possible if g = 3. But 0 fixed points are not possible either for g = 3, as by Theorem 1.2 this would also imply that X is hyperelliptic.

Corollary 3.5. Let X be a curve of genus g 1 mod 4. If Aut(X) has a subgroup H isomorphic to Z/2Z Z/2Z, then X cannot≡ be trigonal. ⊕ Proof. Assume that X is trigonal. Then by Lemma 3.4 each of the 3 involutions in H has exactly 4 fixed points. But then applying the Hurwitz formula to the covering X X/H would give g 3 mod 4. → ≡ The famous Theorem of Faltings says that a curve X of genus g 2 over a number ≥ field K has only finitely many K-rational points. Likewise, it has only finitely many L-rational points for any fixed cubic extension L of K. But since K has infinitely many cubic extensions, X could have infinitely many cubic points in total. Here a cubic point over K means a point P that is L-rational for a cubic extension L of K where L might depend on P . For example, if X is trigonal over K, i.e. if there is a degree 3 cover X Y , → defined over K, such that Y is a projective line over K, then X has infinitely many cubic points over K, namely at least one over each of the infinitely many K-rational points of Y . This is the reason why trigonality is one of the key points to examine, if for a given ensemble of curves one wants to decide which of them have infinitely many

8 cubic points. See for example [JKS1], where we investigated certain modular curves under this aspect to determine which finite groups occur infinitely often as torsion groups of elliptic curves over cubic number fields. Now a curve of genus 2 is always trigonal. If P is any non-Weierstrass point, there is a rational function with a triple pole at P . But the question is whether there is a K-rational trigonal map. By what was just said, a K-rational point that is not a Weierstrass point would suffice for this. But not every genus 2 curve has such a point. In [JKS1, Lemma 2.1] we also showed that if a genus 2 curve has 3 or more K-rational points (Weierstrass or not), then it is trigonal over K, and hence has infinitely many cubic points over K. Considering points that are genuinely cubic over K (i.e. cubic over K but not K-rational), we now prove that a curve of genus 2 either has infinitely many such points or none.

Theorem 3.6. For a curve X of genus 2 over a number ﬁeld K the following are equivalent:

(a) X has a cubic point P over K that is not K-rational.

(b) X has a trigonal map that is deﬁned over K.

Proof. Let L be the cubic extension of K generated by P . Let ι1, ι2, ι3 be the three diﬀerent K-embeddings of L into C. If there is a K-rational function on X with pole divisor ιi(P )+ιj(P )(i = j), this divisor must lie in the canonical class. We can apply a suitable automorphism6 from Gal(K/K) that permutes the three embeddings cyclically, and get that ιj(P )+ιk(P ) also lies in the canonical class. But then ιi(P ) ιk(P ) would be a principal divisor, which is impossible on a curve of positive genus.− Thus the Riemann-Roch space of the K-rational divisor D = ι1(P )+ι2(P )+ι3(P ) contains a K-rational function whose pole divisor actually is D. This function is a K-rational trigonal map. The conclusion (b) (c) is clear, and (c) (a) is trivial. → → Finally we show that among the genus 2 curves that have no Q-rational points some are trigonal over Q and some are not.

Example 3.7. The genus 2 curve

y2 = x6 1 − − has no Q-rational points and no cubic points over Q. In fact, it has no real points at all. The points lying over the point at inﬁnity of the x-line are not real, as 1 is a square in their residue ﬁeld. −

9 Admittedly, the degree 3 map from the curve to C : y2 = x2 1 is deﬁned over − − Q. But C is a genus 0 curve without rational points. So over Q this map is not a trigonal map. Over Q(i), where C is a projective line, the map will of course be trigonal.

Example 3.8. The genus 2 curve

y2 = x6 + 13 − also has no Q-rational points. Here we are using that the elliptic curve y2 = x3 +13 has rank 0 and trivial torsion, and that the points at inﬁnity are not Q-rational for exactly the same reasons as in the previous example. Again, there is an obvious map of degree 3, deﬁned over Q, to the genus 0 curve y2 = x2 + 13. But this curve has Q-rational points, for example ( 2, 3) or ( 3, −2), so it is a projective line over Q, and the degree 3 map to it is± a trigonal± ± ± 3 3 map over Q. Indeed, ( √2, 3) and ( √3, 2) are genuine cubic points on our genus 2 curve. ± ± ± ±

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